#set document(title: "9.4 Two Variance or Standard Deviation F-Test", author: "Rachel Webb") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 9.4#h(0.6em)Two Variance or Standard Deviation F-Test === 9.5.1 The F-Distribution An F-distribution is another special type of distribution for a continuous random variable. Properties of the F-distribution density curve: - Right skewed. - F-scores cannot be negative. - The spread of an F-distribution is determined by the degrees of freedom of the numerator, and by the degrees of freedom of the denominator. The #emph[df] are usually determined by the sample sizes of the two populations or number of groups. - The total area under the curve is equal to 1 or 100%. The shape of the distribution curve changes when the degrees of freedom change. Figure 9-9 shows examples of F-distributions with different degrees of freedom. #figure(figph[Four overlaid F-distribution curves showing how the shape changes with degrees of freedom: df1 = 2, df2 = 27 in red decreases steadily, while df1 = 4, df2 = 7 in dark red, df1 = 14, df2 = 17 in magenta, and df1 = 24, df2 = 27 in blue form progressively taller right-skewed humps, all starting at zero.], alt: "Four overlaid F-distribution curves showing how the shape changes with degrees of freedom: df1 = 2, df2 = 27 in red decreases steadily, while df1 = 4, df2 = 7 in dark red, df1 = 14, df2 = 17 in magenta, and df1 = 24, df2 = 27 in blue form progressively taller right-skewed humps, all starting at zero.", caption: none) Figure 9-9 We will use the F-distribution in several types of hypothesis testing. For now, we are just learning how to find the critical value and probability using the F-distribution. Use the TI-89 Distribution menu; or in Excel F.INV to find the critical values for the F-distribution for tail areas only, depending on the degrees of freedom. When finding a probability given an F-score, use the calculator Fcdf function under the DISTR menu or in Excel use F.DIST. Note that the TI-83 and TI-84 do not come with the INVF function, but you may be able to find the program online or from your instructor. Alternatively, use the calculator at #link("https://homepage.divms.uiowa.edu/~mbognar/applets/f.html")[https://homepage.divms.uiowa.edu/~mbognar/applets/f.html] which will also graph the distribution for you and shade in one tail at a time. You will see the shape of the F-distribution change in the following examples depending on the degrees of freedom used. For your own sketch just make sure you have a positively skewed distribution starting at zero. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The invF your TI-84 does not have] These open the F-distribution the book's three examples use -- the two-tailed pair on df1 = 6, df2 = 14, and the right-tail probability on df1 = 3, df2 = 18 -- each shaded and computed on load, no external applet and no F.INV spreadsheet. For the right-tailed critical value change df to 2 and 12 and set area = 0.95 to get 3.8853. - F(0.025) = F.INV(0.025,6,14) = 0.1888 - F(0.975) = F.INV(0.975,6,14) = 3.5014 - P(F \> 3.894), df1 = 3, df2 = 18 = 0.0263 ] The critical values F#sub[#math.equation(block: false, alt: "α")[$α$]]#sub[/2] and F#sub[1–#math.equation(block: false, alt: "α")[$α$]/2] are for a two-tailed test on the F-distribution curve with area 1 – #math.equation(block: false, alt: "α")[$α$] between the critical values as shown in Figure 9-10. Note that the distribution starts at zero, is positively skewed, and never has negative F-scores. #figure(figph[F-distribution curve, starting at zero and positively skewed, with both tails shaded green: a left tail of area α/2 below the critical value F α/2 and a right tail of area α/2 above F 1-α/2, with the central area labeled 1 - α/2.], alt: "F-distribution curve, starting at zero and positively skewed, with both tails shaded green: a left tail of area α/2 below the critical value F α/2 and a right tail of area α/2 above F 1-α/2, with the central area labeled 1 - α/2.", caption: none) Figure 9-10 #examplebox("Example 1")[][ Compute the critical values F#sub[#math.equation(block: false, alt: "α")[$α$]/2] and F#sub[1–#math.equation(block: false, alt: "α")[$α$]/2] with #emph[df]#sub[1] = 6 and #emph[df]#sub[2] = 14 for a two-tailed test, #math.equation(block: false, alt: "α")[$α$] = 0.05. #solutionbox[ Start by drawing the curve and finding the area in each tail. For this case, it would be an area of #math.equation(block: false, alt: "α")[$α$]/2 in each tail. Then use technology to find the F-scores. Most technology only asks for the area to the left of the F-score you are trying to find. In Excel the function for F#sub[#math.equation(block: false, alt: "α")[$α$]/2] is F.INV(area in left-tail,#emph[df]#sub[1],#emph[df]#sub[2]). There is only one function, so use areas 0.025 and 0.975 in the left tail. For this example, we would have critical values F#sub[0.025] = F.INV(0.025,6,14) = 0.1888 and F#sub[0.975] = F.INV(0.975,6,14) = 3.5014. See Figure 9-11. #figure(figph[Positively skewed F-distribution curve with tails of area α/2 shaded below the critical value 0.1888 and above 3.5014, and the central area labeled 1 - α/2.], alt: "Positively skewed F-distribution curve with tails of area α/2 shaded below the critical value 0.1888 and above 3.5014, and the central area labeled 1 - α/2.", caption: none) Figure 9-11 We have to calculate two distinct F-scores unlike symmetric distribution where we could just do ±z-score or ±t-score. Note if you were doing a one-tailed test then do not divide alpha by two and use area = #math.equation(block: false, alt: "α")[$α$] for a left-tailed test and area = 1 – #math.equation(block: false, alt: "α")[$α$] for a right-tailed test. ] ] #examplebox("Example 2")[][ Find the critical value for a right-tailed test with denominator degrees of freedom of 12 and numerator degrees of freedom of 2 with a 5% level of significance. #solutionbox[ Draw the curve and shade in the top 5% of the upper tail since #math.equation(block: false, alt: "α")[$α$] = 0.05, see Figure 9-12. When using technology, you will need the area to the left of the critical value that you are trying to find. This would be 1 – #math.equation(block: false, alt: "α")[$α$] = 0.95. Then identify the degrees of freedom. The first degrees of freedom are the numerator #emph[df], therefore #emph[df]#sub[1] = 2. The second degrees of freedom are the denominator #emph[df], therefore #emph[df]#sub[2] = 12. Using Excel, we would have =F.INV(0.95,2,12) = 3.8853. #figure(figph[Steeply decreasing F-distribution curve with the area 1 - α = 0.95 under the body and the right tail of area α = 0.05 shaded beyond the critical value F 1-α = 3.8853.], alt: "Steeply decreasing F-distribution curve with the area 1 - α = 0.95 under the body and the right tail of area α = 0.05 shaded beyond the critical value F 1-α = 3.8853.", caption: none) Figure 9-12 ] ] #examplebox("Example 3")[][ Compute P(F \> 3.894), with #emph[df]#sub[1] = 3 and #emph[df]#sub[2] = 18 #solutionbox[ In Excel, use the function F.DIST(x,deg\_freedom1,deg\_freedom2,cumulative). Always use TRUE for the cumulative. The F.DIST function will find the probability (area) below F. Since we want the area above F we would need to also use the complement rule. The formula would be =1-F.DIST(3.894,3,18,TRUE) = 0.0263. TI-84: The TI-84 calculator has a built in F-distribution. Press \[2#super[nd]\] \[DISTR\] (this is F5: DISTR in the STAT app in the TI-89), then arrow down until you get to the Fcdf and press \[Enter\]. Depending on your calculator, you may not get a prompt for the boundaries and df. If you just see Fcdf( then you will need to enter each the lower boundary, upper boundary, #emph[df]#sub[1], and #emph[df]#sub[2] with a comma between each argument. The lower boundary is the 3.394 and the upper boundary is infinity (TI-83 and 84 use a really large number instead of ∞), then enter the two degrees of freedom. Press \[Paste\] and then \[Enter\], this will put the Fcdf(3.894,1E99,3,18) on your screen and then press \[Enter\] again to calculate the value. #figure(figph[TI-84 screens computing an F-distribution tail area: the DISTR menu highlights Fcdf, the input screen shows lower: 3.894, upper: 1E99, dfNumer: 3, dfDenom: 18 with Paste, and the home screen shows Fcdf(3.894,1E99,3,18) returning 0.0262987803.], alt: "TI-84 screens computing an F-distribution tail area: the DISTR menu highlights Fcdf, the input screen shows lower: 3.894, upper: 1E99, dfNumer: 3, dfDenom: 18 with Paste, and the home screen shows Fcdf(3.894,1E99,3,18) returning 0.0262987803.", caption: none) Figure 9-13. #figure(figph[Interactive F-distribution calculator for X ~ F(df1, df2) with df1 = 3, df2 = 18, and x = 3.894, returning P(X \> x) = 0.0263; below is a right-skewed F density curve on an x-axis from about 0.2 to 5, with μ = E(X) = 1.125, σ = SD(X) = 1.07, and σ² = Var(X) = 1.145 reported underneath.], alt: "Interactive F-distribution calculator for X ~ F(df1, df2) with df1 = 3, df2 = 18, and x = 3.894, returning P(X > x) = 0.0263; below is a right-skewed F density curve on an x-axis from about 0.2 to 5, with μ = E(X) = 1.125, σ = SD(X) = 1.07, and σ² = Var(X) = 1.145 reported underneath.", caption: none) Figure 9-13 ] ] === 9.5.2 Hypothesis Test for Two Variances Sometimes we will need to compare the variation or standard deviation between two groups. For example, let’s say that the average delivery time for two locations of the same company is the same but we hear complaint of inconsistent delivery times for one location. We can use an F-test to see if the standard deviations for the two locations was different. There are three types of hypothesis tests for comparing the ratio of two population variances , see Figure 9-14. #figure(figph[Three-column table of hypothesis tests for two variances. Two-tailed: H0: σ1² = σ2², H1: σ1² ≠ σ2², with both tails of a right-skewed F curve shaded green. Right-tailed: H1: σ1² \> σ2², right tail shaded. Left-tailed: H1: σ1² \< σ2², left tail shaded. A bottom row restates each pair as ratios, such as H0: σ1²/σ2² = 1 versus H1: σ1²/σ2² \> 1.], alt: "Three-column table of hypothesis tests for two variances. Two-tailed: H0: σ1² = σ2², H1: σ1² ≠ σ2², with both tails of a right-skewed F curve shaded green. Right-tailed: H1: σ1² > σ2², right tail shaded. Left-tailed: H1: σ1² < σ2², left tail shaded. A bottom row restates each pair as ratios, such as H0: σ1²/σ2² = 1 versus H1: σ1²/σ2² > 1.", caption: none) Figure 9-14 If we take the square root of the variance, we get a standard deviation. Therefore, taking the square root of both sides of the hypotheses, we can also use the same test for standard deviations. We use the following notation for the hypotheses. There are 3 types of hypothesis tests for comparing the population standard deviations σ#sub[1]#sub[/]σ#sub[2], see Figure 9-15. #figure(figph[Three-column table of hypothesis tests for two standard deviations. Two-tailed: H0: σ1 = σ2, H1: σ1 ≠ σ2, with both tails of a right-skewed F curve shaded green. Right-tailed: H1: σ1 \> σ2, right tail shaded. Left-tailed: H1: σ1 \< σ2, left tail shaded.], alt: "Three-column table of hypothesis tests for two standard deviations. Two-tailed: H0: σ1 = σ2, H1: σ1 ≠ σ2, with both tails of a right-skewed F curve shaded green. Right-tailed: H1: σ1 > σ2, right tail shaded. Left-tailed: H1: σ1 < σ2, left tail shaded.", caption: none) Figure 9-15 #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The F-test is a statistical test for comparing the variances or standard deviations from two populations. The formula for the test statistic is #math.equation(block: false, alt: "F equals the fraction s sub 1 squared over s sub 2 squared")[$F = frac(s_(1)^(2), s_(2)^(2))$]. With numerator degrees of freedom = N#emph[df]= n#sub[1] – 1, and denominator degrees of freedom = D#emph[df] = n#sub[2] – 1. This test may only be used when both populations are independent and normally distributed. ] Important: This F-test is #strong[not]robust (a statistic is called “robust” if it still performs reasonably well even when the necessary conditions are not met). In particular, this F-test demands that both populations be normally distributed even for larger sample sizes. This F-test yields unreliable results when this condition is not met. \# "Not robust" has a price tag. Here it is. \# Both samples come from the SAME skewed population, so the variances really ARE \# equal: a 5% test should reject H0 about 5% of the time. Watch what happens. set.seed(9) n1 \<- 20; n2 \<- 18 rate \<- function(gen) mean(replicate(5000, { x \<- gen(n1); y \<- gen(n2) Fstat \<- var(x) / var(y) tail \<- pf(Fstat, n1 - 1, n2 - 1) 2 \* min(tail, 1 - tail) \< 0.05 \# two-tailed, alpha = 0.05 })) rate(rexp) \# skewed population -\> about 0.26, five times the 5% you asked for rate(rnorm) \# normal population -\> back near 0.05, as advertised \# Change rexp to runif (light tails) to see the error rate miss the other way. The traditional method (or critical value method), and the p-value method are performed with steps that are identical to those when performing hypothesis tests from previous sections. #examplebox("Example 4")[][ A researcher claims that IQ scores of university students vary less than (have a smaller variance than) IQ scores of community college students. Based on a sample of 28 university students, the sample standard deviation 10, and for a sample of 25 community college students, the sample standard deviation 12. Test the claim using the traditional method of hypothesis testing with a level of significance #math.equation(block: false, alt: "α")[$α$] = 0.05. Assume that IQ scores are normally distributed. #solutionbox[ 1. The claim is “IQ scores of university students (Group 1) have a smaller variance than IQ scores of community college students (Group 2).” This is a left-tailed test; therefore, the hypotheses are: \\(\\begin{aligned} #linebreak() &H\_{0}: \\sigma\_{1}^{2}=\\sigma\_{2}^{2} \\\\ #linebreak() &H\_{1}: \\sigma\_{1}^{2}\<\\sigma\_{2}^{2} #linebreak() \\end{aligned}\\). 2. We are using the F-test because we are performing a test about two population variances. We can use the F-test only if we assume that both populations are normally distributed. We will assume that the selection of each of the student groups was independent. The problem gives us s#sub[1] = 10, n#sub[1] = 28, s#sub[2] = 12, and n#sub[2] = 25. The formula for the test statistic is #math.equation(block: false, alt: "F equals the fraction s sub 1 squared over s sub 2 squared equals the fraction 10 squared over 12 squared equals 0.6944")[$F = frac(s_(1)^(2), s_(2)^(2)) = frac(10^(2), 12^(2)) = 0.6944$]. 3. The critical value for a left-tailed test with a level of significance #math.equation(block: false, alt: "α")[$α$] = 0.05 is found using the invF program or Excel. See Figure 9-16. Using Excel: The critical value is F#sub[#math.equation(block: false, alt: "α")[$α$]] =F.INV(0.05,27,24) = 0.5182. #figure(figph[Right-skewed F-distribution curve with the left-tail area α = 0.05 shaded green and the remaining area labeled 1 - α = 0.95; below the axis the test statistic F = 0.6944 and the critical value Fα = 0.5182 are marked.], alt: "Right-skewed F-distribution curve with the left-tail area α = 0.05 shaded green and the remaining area labeled 1 - α = 0.95; below the axis the test statistic F = 0.6944 and the critical value Fα = 0.5182 are marked.", caption: none) Figure 9-16 4. Decision: Compare the test statistic F = 0.6944 with the critical value F#sub[#math.equation(block: false, alt: "α")[$α$]] = 0.5182, see Figure 9-16. Since the test statistic is not in the rejection region, we do not reject H#sub[0]. 5. Summary: There is not enough evidence to support the claim that the IQ scores of university students have a smaller variance than IQ scores of community college students. ] ] #examplebox("Example 5")[][ A random sample of 20 graduate college students and 18 undergraduate college students indicated these results concerning the amount of time spent in volunteer service per week. At #math.equation(block: false, alt: "α")[$α$] = 0.01 level of significance, is there sufficient evidence to conclude that graduate students have a higher standard deviation of the number of volunteer hours per week compared to undergraduate students? Assume that number of volunteer hours per week is normally distributed. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Same F-test, both methods, right degrees of freedom] F = 3.5/2.2 = 1.5909 with numerator df = n1 - 1 = 19 and denominator df = n2 - 1 = 17. The first link is the book's F.INV.RT(0.01,19,17) critical value 3.1857, the second its F.DIST.RT p-value 0.1704 -- both say do not reject at alpha = 0.01. The third link is the Sacramento/Portland variance test, where F \< 1 so you read the LEFT tail and double it. Try swapping df and df2 to see why the order of the groups matters here. - critical value F.INV.RT(0.01,19,17) = 3.1857 - p-value F.DIST.RT(1.5909,19,17) = 0.1704 - P(F \< 0.82796), df1 = 16, df2 = 15 = 0.3553, doubled = 0.7106 ] Graduate Undergraduate Sample Mean 3.8 2.5 Sample Variance 3.5 2.2 Sample Size 20 18 #solutionbox[ Assumptions: The two populations we are comparing are graduate and undergraduate college students. We are given that the number of volunteer hours per week is normally distributed. We are told that the samples were randomly selected and should therefore be independent. Using the Traditional Method 1. We are trying to determine whether the standard deviation of the number of volunteer hours per week for graduate students (Group 1) is larger than undergraduate students (Group 2) or σ#sub[1] \> σ#sub[2]. Therefore, the hypotheses are: \\(\\begin{aligned} #linebreak() &\\mathrm{H}\_{0}: \\sigma\_{1}=\\sigma\_{2} \\\\ #linebreak() &\\mathrm{H}\_{1}: \\sigma\_{1}\>\\sigma\_{2} #linebreak() \\end{aligned}\\) 2. We are given that #math.equation(block: false, alt: "s sub 1 squared")[$s_(1)^(2)$] =3.5, #math.equation(block: false, alt: "s sub 2 squared")[$s_(2)^(2)$] = 2.3, n#sub[1] =20 and n#sub[2] =18. Note variances were given, so do not square the numbers again. The test statistic is, #math.equation(block: false, alt: "F equals the fraction s sub 1 squared over s sub 2 squared equals the fraction 3.5 over 2.2 equals 1.5909")[$F = frac(s_(1)^(2), s_(2)^(2)) = frac(3.5, 2.2) = 1.5909$]. 3. Draw and label the distribution with the critical value for a right-tailed F-test with numerator degrees of freedom = n#sub[1] – 1 = 19, and with denominator degrees of freedom = n#sub[2] – 1 = 17. See Figure 9-17. Use right-tail area #math.equation(block: false, alt: "α")[$α$] = 0.01 in Excel F#sub[1–#math.equation(block: false, alt: "α")[$α$]] =F.INV.RT(0.01,19,17) to find the critical value 3.1857. #figure(figph[Right-skewed F-distribution curve with the right-tail area α = 0.01 shaded green beyond the critical value Fα = 3.1857, the body labeled 1 - α = 0.99, and the test statistic F = 1.5909 marked to the left of the shaded region.], alt: "Right-skewed F-distribution curve with the right-tail area α = 0.01 shaded green beyond the critical value Fα = 3.1857, the body labeled 1 - α = 0.99, and the test statistic F = 1.5909 marked to the left of the shaded region.", caption: none) Figure 9-17 4. Decision: Since the test statistic is not in the rejection region, we do not reject H#sub[0]. 5. Summary: There is not enough evidence to support the claim that the population standard deviation of the number of volunteer hours per week for graduate college students is higher than undergraduate college students. #strong[Using the p-value method] 1. Step 1 remains the same. Therefore, the hypotheses are: \\(\\begin{aligned} #linebreak() &\\mathrm{H}\_{0}: \\sigma\_{1}=\\sigma\_{2} \\\\ #linebreak() &\\mathrm{H}\_{1}: \\sigma\_{1}\>\\sigma\_{2} #linebreak() \\end{aligned}\\) 2. Step 2 remains the same. The test statistic is, #math.equation(block: false, alt: "F equals the fraction s sub 1 squared over s sub 2 squared equals the fraction 3.5 over 2.2 equals 1.5909")[$F = frac(s_(1)^(2), s_(2)^(2)) = frac(3.5, 2.2) = 1.5909$]. 3. Compute the p-value using either the Fcdf on the calculator or Excel. If your test statistic is less than 1, then find the area to the left of the test statistic, if F is above 1 then find the area to the right of the test statistic. If you have a two-tailed test then double your tail area. #figure(figph[TI-84 screen computing 3.5/2.2 = 1.590909091, then Fcdf(1.5909,1E9… with the remaining arguments running off the screen, returning 0.1703622488.], alt: "TI-84 screen computing 3.5/2.2 = 1.590909091, then Fcdf(1.5909,1E9… with the remaining arguments running off the screen, returning 0.1703622488.", caption: none) TI: Fcdf(lower,upper,#emph[df]#sub[1],#emph[df]#sub[2]) = Fcdf(1.5909,1E99,19,17). Excel: =F.DIST.RT(1.5909,19,17) = 0.1704. 4. Decision: Since the p-value = 0.1704 is greater than #math.equation(block: false, alt: "α")[$α$] = 0.01, we “Do Not Reject H#sub[0].” 5. Step 5, the summary remains the same. There is not enough evidence to support the claim that the population standard deviation of the number of volunteer hours per week for graduate college students is higher than undergraduate college students. Alternatively use the following 2-Sample F-test shortcut on the TI calculator. #strong[TI-84:] Press the \[STAT\] key, arrow over to the \[TESTS\] menu, arrow down to the option \[E:2-SampFTest\] and press the \[ENTER\] key. Arrow over to the \[Stats\] menu and press the \[Enter\] key. Then type in the s#sub[1], n#sub[1], s#sub[2], n#sub[2], arrow over to the #math.equation(block: false, alt: "not equal to")[$≠$], \<, \> sign that is the same in the problem’s alternative hypothesis statement, then press the \[ENTER\] key, arrow down to \[Calculate\] and press the \[ENTER\] key. The calculator returns the test statistic F and the p-value. #figure(figph[Three TI-84 screens for a 2-SampFTest: the STAT TESTS menu with E:2-SampFTest highlighted; the input screen with Stats selected, Sx1: 1.87082869, n1: 20, Sx2: √(2.2), n2: 18, and alternative σ1 \> σ2 chosen; and the results screen showing F = 1.590909091, p = 0.1703592883, Sx1 = 1.87082869, Sx2 = 1.4832397, n1 = 20.], alt: "Three TI-84 screens for a 2-SampFTest: the STAT TESTS menu with E:2-SampFTest highlighted; the input screen with Stats selected, Sx1: 1.87082869, n1: 20, Sx2: √(2.2), n2: 18, and alternative σ1 > σ2 chosen; and the results screen showing F = 1.590909091, p = 0.1703592883, Sx1 = 1.87082869, Sx2 = 1.4832397, n1 = 20.", caption: none) #strong[Note: You have to put the standard deviation in the calculator, not the variance.] #strong[TI-89:] Go to the \[Apps\]#strong[Stat/List Editor], then push 2#super[nd] then F6 \[Tests\], then select #strong[9: 2-]#strong[SampFTest]. Then type in the s#sub[1], n#sub[1], s#sub[2], n#sub[2] (or list names list1 & list2), select the sign #math.equation(block: false, alt: "not equal to")[$≠$], \<, \> that is the same in the problem’s alternative hypothesis statement, press the \[ENTER\] key to calculate. The calculator returns the F-test statistic and the p-value. ] ] #examplebox("Example 6")[][ A researcher is studying the variability in electricity (in kilowatt hours) people from two different cities use in their homes. Random samples of 17 days in Sacramento and 16 days in Portland are given below. Test to see if there is a difference in the variance of electricity use between the two cities at α = 0.10. Assume that electricity use is normally distributed, use the p-value method. #figure(figph[Two data tables of daily electricity use in kilowatt hours. Sacramento, 17 values: 474, 414, 692, 467, 443, 605, 419, 277, 670, 696, 783, 577, 813, 694, 565, 663, 884. Portland, 16 values: 783, 587, 527, 546, 442, 107, 728, 662, 371, 427, 277, 474, 605, 293, 320, 555.], alt: "Two data tables of daily electricity use in kilowatt hours. Sacramento, 17 values: 474, 414, 692, 467, 443, 605, 419, 277, 670, 696, 783, 577, 813, 694, 565, 663, 884. Portland, 16 values: 783, 587, 527, 546, 442, 107, 728, 662, 371, 427, 277, 474, 605, 293, 320, 555.", caption: none) #solutionbox[ The populations are independent and normally distributed. The hypotheses are \\(\\begin{aligned} #linebreak() &\\mathrm{H}\_{0}: \\sigma\_{1}^{2}=\\sigma\_{2}^{2} \\\\ #linebreak() &\\mathrm{H}\_{1}: \\sigma\_{1}^{2} \\neq \\sigma\_{2}^{2} #linebreak() \\end{aligned}\\) Use technology to compute the standard deviations and sample sizes. Enter the Sacramento data into list 1, then do 1-Var Stats L1 and you should get s#sub[1] = 163.2362 and n#sub[1] = 17. Enter the Portland data into list 2, then do 1-Var Stats L2 and you should get s#sub[2] = 179.3957 and n#sub[2] = 16. Alternatively, use Excel’s descriptive statistics. The test statistic is The p-value would be double the area to the left of F = 0.82796 (Use double the area to the right if the test statistic is \> 1). #figure(figph[TI-84 screen computing 163.2362²/179.39… (the expression runs off the screen) = 0.82795912, then Fcdf(0,.82796,1… returning 0.3552760097, then Ans\*2 = 0.7105520193, the two-tailed p-value.], alt: "TI-84 screen computing 163.2362²/179.39… (the expression runs off the screen) = 0.82795912, then Fcdf(0,.82796,1… returning 0.3552760097, then Ans*2 = 0.7105520193, the two-tailed p-value.", caption: none) Using the TI calculator Fcdf(0,0.82796,16,15). In Excel we get the p-value =2\*F.DIST(E8,E7,F7,TRUE) = 0.7106. Since the p-value is greater than alpha, we would fail to reject H#sub[0]. There is no statistically significant difference between variance of electricity use between Sacramento and Portland. #strong[Excel:] When you have raw data, you can use Excel to find all this information using the Data Analysis tool. Enter the data into Excel, then choose Data \> Data Analysis \> F-Test: Two Sample for Variances. #figure(figph[Excel worksheet with Sacramento data in column A and Portland data in column B, and the Data Analysis dialog open with F-Test Two-Sample for Variances highlighted in the Analysis Tools list, with OK, Cancel, and Help buttons.], alt: "Excel worksheet with Sacramento data in column A and Portland data in column B, and the Data Analysis dialog open with F-Test Two-Sample for Variances highlighted in the Analysis Tools list, with OK, Cancel, and Help buttons.", caption: none) Enter the necessary information as we did in previous sections (see below) and select OK. Note that Excel only does a one-tail F-test so use #math.equation(block: false, alt: "α")[$α$]/2 = 0.10/2 = 0.05 in the Alpha box. #figure(figph[Excel F-Test Two-Sample for Variances dialog with Variable 1 Range \$A\$1:\$A\$18, Variable 2 Range \$B\$1:\$B\$17, the Labels box checked, Alpha 0.05, and Output Range \$D\$1 selected.], alt: "Excel F-Test Two-Sample for Variances dialog with Variable 1 Range $A$1:$A$18, Variable 2 Range $B$1:$B$17, the Labels box checked, Alpha 0.05, and Output Range $D$1 selected.", caption: none) We get the following output. Note you can only use the critical value in Excel for a left-tail test. #figure(figph[Excel F-Test Two-Sample for Variances output for Sacramento and Portland: means 596.2352941 and 481.5, variances 26646.06618 and 32182.8, observations 17 and 16, df 16 and 15, F = 0.827959847, P(F\<=f) one-tail = 0.355275877, and F Critical one-tail = 0.425129803, with a notes column explaining each row, such as p-value for a left-tailed test.], alt: "Excel F-Test Two-Sample for Variances output for Sacramento and Portland: means 596.2352941 and 481.5, variances 26646.06618 and 32182.8, observations 17 and 16, df 16 and 15, F = 0.827959847, P(F<=f) one-tail = 0.355275877, and F Critical one-tail = 0.425129803, with a notes column explaining each row, such as p-value for a left-tailed test.", caption: none) Excel for some reason only does the smaller tail area for the F-test, so you will need to double the p-value for a two-tailed test, p-value = 0.355275877\*2 = 0.7106. ] ]